Here is another question though:
Find the derivative of arctan(x), using arctan(tan(x)) = x.
This also has me stumped.

[itex]g'(x) = \arctan(x)[/itex]

I know that:
[itex]f'(x) = f'[g(x)] * g'(x)[/itex]

But, I don't see what I can do with arctan(x), except for the obvious:

[itex]\arctan(\arctan(\tan(x)))[/itex]
which doesn't help at all.

How would I proceed? I'm suppoed to use the derivative of a "functionsfunction", not sure what it's called in english.

Say y = arctan(x) then tan(y) = tan(atan(x)). But tan(atan(x))=x for all x. So we have tan(y) = x deriving with respect to x on left side and right side we get:
y' * (tan(y)^2 + 1) = 1 => y' = 1/(tan(y)^2+1). But what is tan(y)? Well tan(y)=x so the derivate of arctan(x) is 1/(x^2+1).

Say y = arctan(x) then tan(y) = tan(atan(x)). But tan(atan(x))=x for all x. So we have tan(y) = x deriving with respect to x on left side and right side we get:
y' * (tan(y)^2 + 1) = 1 => y' = 1/(tan(y)^2+1). But what is tan(y)? Well tan(y)=x so the derivate of arctan(x) is 1/(x^2+1).

I'm not understanding this :(
How do you get [itex]\frac{1}{\cos^2(x) + 1}[/itex]
?

[tex]\tan (\arctan (x))=x[/tex]
[tex]\frac{d}{dx} \tan (\arctan (x)) = \frac{d}{dx} x[/tex]
[tex]\arctan'(x) \times \sec^2(\arctan(x))=1[/tex]
[tex]\arctan'(x) = \cos^2(\arctan(x))[/tex]
Now
[tex]\cos(\arctan(x))=\frac{\pm 1}{\sqrt{1-x^2}}[/tex]
(Consider, for example, a right triangle where the adjacent side is 1, the opposite side is [tex]x[/tex] and the hypotenuse is [tex]\sqrt{1+x^2}[/tex])
So we can substitute that in:
[tex]\arctan'(x)=(\frac{\pm 1}{\sqrt{1-x^2}})^2=\frac{1}{1+x^2}[/tex]
the [tex]\pm[/tex] drops out because of the square.